Problem 36 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ \left(\frac{x+1}{x}\right)^{2}+4\left(\frac{x+1}{x}\right)+3=0 $$

Step-by-Step Solution

Answer

The real solutions are \( x = -\frac{1}{2} \) and \( x = -\frac{1}{4} \).

Step 1: Substitute the Variable

Let \( u = \frac{x+1}{x} \). This simplifies the original equation into a quadratic equation: \( u^2 + 4u + 3 = 0 \).

Step 2: Solve the Quadratic Equation

Factor the quadratic equation \( u^2 + 4u + 3 = 0 \) into \( (u+1)(u+3) = 0 \). This gives the solutions \( u = -1 \) and \( u = -3 \).

Step 3: Substitute Back to Find x

Reverse the substitution by setting \( \frac{x+1}{x} = u \) for each solution. So, for \( u = -1 \), solve \( \frac{x+1}{x} = -1 \), and for \( u = -3 \), solve \( \frac{x+1}{x} = -3 \).

Step 4: Solve for x when u = -1

For \( \frac{x+1}{x} = -1 \), multiply through by \( x \) to get \( x + 1 = -x \). Simplifying gives \( 2x = -1 \) or \( x = -\frac{1}{2} \).

Step 5: Solve for x when u = -3

For \( \frac{x+1}{x} = -3 \), multiply through by \( x \) to get \( x + 1 = -3x \). Simplifying gives \( 4x = -1 \) or \( x = -\frac{1}{4} \).

Step 6: Conclusion

The solutions for \( x \) are \( x = -\frac{1}{2} \) and \( x = -\frac{1}{4} \).

Key Concepts

Substitution Method

When dealing with complex equations, the substitution method is a powerful tool for simplifying and solving them. This technique involves replacing a section of the equation with a single variable. This helps to transform a complicated equation into a simpler, often quadratic form.
In the context of our problem, we introduced a substitution by letting \( u = \frac{x+1}{x} \). This turned the original equation into the much simpler quadratic equation \( u^2 + 4u + 3 = 0 \).
The substitution step is crucial because it allows us to focus on solving a standard quadratic equation instead of directly tackling a more intricate expression that involves fractions. By solving for "\( u \)," we create a path to easily find "\( x \)," the original variable we are interested in.

Factoring Quadratic Equations

Once we have simplified a problem to a quadratic equation, such as \( u^2 + 4u + 3 = 0 \), factoring is a straightforward method for finding solutions. Factoring involves expressing the quadratic as a product of two binomials.
In this example, we factored \( u^2 + 4u + 3 \) into \((u+1)(u+3) = 0\). This shows us that the solutions for "\( u \)" are the values that make each factor zero: \( u+1 = 0 \) and \( u+3 = 0 \). So, \( u = -1 \) and \( u = -3 \).
This technique is valuable because it simplifies solving quadratic equations into just finding numbers that nullify the product, further easing the process of finding the original solution for \( x \).

Solving Rational Equations

After solving for the substituted variable \( u \), the next step is often solving rational equations to find the original variable. Rational equations are fractions involving a variable in the denominator.
We substitute back, using \( \frac{x+1}{x} = u \), to find \( x \). For each value of \( u \), we solve the respective rational equation:

• For \( u = -1 \), we solve \( \frac{x+1}{x} = -1 \) by clearing the fraction, which leads to \( x = -\frac{1}{2} \).
• For \( u = -3 \), we solve \( \frac{x+1}{x} = -3 \) the same way, which gives \( x = -\frac{1}{4} \).

Rational equations can seem intimidating, but by isolating the fraction and simplifying, such as multiplying through by \( x \), you can solve them just like a simple equation. This final step confirms the solution for the original problem in terms of \( x \), concluding the process successfully.